【leetcode】1220. Count Vowels Permutation

题目如下：

Given an integer n, your task is to count how many strings of length n can be formed under the following rules:

Each character is a lower case vowel ('a', 'e', 'i', 'o', 'u')

Each vowel 'a' may only be followed by an 'e'.

Each vowel 'e' may only be followed by an 'a' or an 'i'.

Each vowel 'i' may not be followed by another 'i'.

Each vowel 'o' may only be followed by an 'i' or a 'u'.

Each vowel 'u' may only be followed by an 'a'.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:
Input: n = 1
Output: 5
Explanation: All possible strings are: "a", "e", "i" , "o" and "u".
Example 2:
Input: n = 2
Output: 10
Explanation: All possible strings are: "ae", "ea", "ei", "ia", "ie", "io", "iu", "oi", "ou" and "ua".
Example 3:
Input: n = 5
Output: 68
Constraints:

1 <= n <= 2 * 10^4

解题思路：用动态规划的方法，记dp[i][j] 为第i个位置放第j个元音字母的情况下可以组成的字符串的个数，很容易得到状态转移方程。例如如果第i个元素是'e'，那么第i-1的元素就只能是'a'或者'i'，有 dp[i]['e'] = dp[i-1]['a'] + dp[i-1]['i']，最后只有求出第n个元素放这五个元音字母的个数的和即可。

代码如下：

class Solution(object):
    def countVowelPermutation(self, n):
        """
        :type n: int
        :rtype: int
        """
        dp = [[0] * 5 for _ in range(n)]
        vowels = ['a', 'e', 'i', 'o', 'u']
        dp[0][0] = dp[0][1] =dp[0][2] =dp[0][3] = dp[0][4] = 1
        for i in range(1,len(dp)):
            for j in range(len(vowels)):
                if vowels[j] == 'a':
                    dp[i][j] += dp[i - 1][vowels.index('e')]
                    dp[i][j] += dp[i - 1][vowels.index('u')]
                    dp[i][j] += dp[i - 1][vowels.index('i')]
                elif vowels[j] == 'e':
                    dp[i][j] += dp[i - 1][vowels.index('a')]
                    dp[i][j] += dp[i - 1][vowels.index('i')]
                elif vowels[j] == 'i':
                    dp[i][j] += dp[i - 1][vowels.index('e')]
                    dp[i][j] += dp[i - 1][vowels.index('o')]
                elif vowels[j] == 'o':
                    dp[i][j] += dp[i - 1][vowels.index('i')]
                elif vowels[j] == 'u':
                    dp[i][j] += dp[i - 1][vowels.index('i')]
                    dp[i][j] += dp[i - 1][vowels.index('o')]
        return sum(dp[-1]) % (10**9 + 7)

原文地址：https://www.cnblogs.com/seyjs/p/11646665.html